# Proof of Baire’s category theorem

The book “Probability with martingales” by David Williams provides a proof of Baire’s category theorem in p. 203-204. This blog post reproduces the proof in greater detail to clarify some of the statements that are left to the book reader to confirm.

Baire’s category theorem is stated as follows:

If a complete metric space $S$ can be written as a union $S=\underset{n\in\mathbb{N}}{\cup}F_n$ of a countable sequence of closed sets $F_n,n\in\mathbb{N}$, then at least one of these closed sets contains an open ball.

The theorem was proved for $\mathbb{R}^d$ by René-Louis Baire in his PhD thesis “Sur les fonctions de variable réelles“.

Assume a complete metric space $(S,\rho)$ with some metric $\rho:S\times S\rightarrow \mathbb{R}$, which need not be necessarily the Euclidean metric space $(\mathbb{R}^d, |\cdot|)$.

Start by assuming that none of $F_n,n\in\mathbb{N}$, contains an open ball. Moreover, assume that none of the complements $F_n^c,n\in\mathbb{N}$, is empty. If there exists an empty complement $F_n^c=\emptyset$, then $F_n=S$, which is a trivial case of finite union that will be treated at the end of the proof separately.

Since $F_1$ is closed, $F_1^c$ is open, i.e. $F_1^c=(F_1^c)^{\mathrm{o}}$. Due to $F_1^c$ being non-empty, there exists $x_1\in F_1^c=(F_1^c)^{\mathrm{o}}$. Hence, there exists $\epsilon_1 >0$ so that the open ball $B(x_1,\epsilon_1)$ with center $x_1$ and radius $\epsilon_1$ satisfies $B(x_1,\epsilon_1)\subseteq F_1^c$.

By assumption, $F_2$ contains no open ball, so $B(x_1, \epsilon_1/2)\not\subseteq F_2$, which means that $B(x_1, \epsilon_1/2)\cap F_2^c \neq\emptyset$. Furthermore, $B(x_1, \epsilon_1/2)\cap F_2^c$ is an open set as the intersection of the open sets $B(x_1, \epsilon_1/2)$ and $F_2^c$. Thus, there exist $x_2\in B(x_1, \epsilon_1/2)\cap F_2^c$ and $\epsilon_2>0$ with $\epsilon_2<\min \{\epsilon_1/2,1/2\}$ such that $B(x_2,\epsilon_2)\subseteq B(x_1,\epsilon_1/2)\cap F_2^c$. Note that $B(x_2,\epsilon_2)\subseteq B(x_1,\epsilon_1/2)$.

Inductively, choose $\epsilon_n<\min \{\epsilon_{n-1}/2,1/n\},n\in\mathbb{N}$, to introduce the n-th ball $B(x_n,\epsilon_n)\subseteq B(x_{n-1},\epsilon_{n-1}/2)\cap F_n^c$. This way, a countable sequence of nested balls is defined with $B(x_n,\epsilon_n)\subseteq B(x_{n-1},\epsilon_{n-1}/2)$. It will be shown that the centers of these balls make up a Cauchy sequence $(x_n)$. According to the Archimedean property, for any $\epsilon > 0$, there exists $n_0\in\mathbb{N}$ such that $0< 1/n_o < \epsilon$. Let $m,n\in\mathbb{N}$ be natural numbers with $m>n_o$ and $n>n_o$. Using the triangle inequality, $\rho (x_m, x_n) \le \rho (x_m, x_{n_o})+\rho (x_{n_o}, x_n)$. Obviously, $x_n\in B(x_n,\epsilon_n)$. Since $n>n_o$, it follows that $B(x_n,\epsilon_n)\subseteq B(x_{n_o}, \epsilon_{n_o}/2)$, so $x_n\in B(x_{n_o}, \epsilon_{n_o}/2)$. Apparently, $x_n\in B(x_{n_o}, \epsilon_{n_o}/2)$ implies $\rho(x_n,x_{n_o})<\epsilon_{n_o}/2$. As $\epsilon_{n_o}<1/n_o$ and $1/n_o < \epsilon$, it follows that $\rho(x_n,x_{n_o})<\epsilon/2$. Similarly, $\rho(x_m,x_{n_o})<\epsilon/2$, which leads to $\rho (x_m, x_n) < \epsilon$, so the ball centers constitute the Cauchy sequence $(x_n)$. The metric space $(S, \rho)$ has been assumed to be complete, so the Cauchy sequence $(x_n)$ is convergent, therefore its limit $x:=\lim{x_n}$ exists in $S$.

The next step is to show that the limit $x$ of the ball centers $(x_n)$ belongs to the intersection of these balls, i.e. $x\in\underset{n\in\mathbb{N}}{\cap}B(x_n,\epsilon_n)$. Let $n\in \mathbb{N}$ and $m\in\mathbb{N}$ with $m>n$. By applying the triangle inequality, $\rho (x, x_n) \le \rho (x, x_m)+\rho (x_m, x_n)$. Notice that $m>n$ yields $x_m\in B(x_m,\epsilon_m)\subseteq B(x_n, \epsilon_n/2)$, hence $\rho (x_m, x_n) <\epsilon_n/2$, which leads to $\rho (x, x_n) \le \rho (x, x_m)+\epsilon_n/2$. Observe that $(x_m)$ is a subsequence of $(x_n)$, which means that $\underset{m\rightarrow\infty}{\lim}\rho (x, x_m)=0$. So, taking the limit as $m\rightarrow \infty$ produces  $\rho (x, x_n) \le \epsilon_n/2$, whence $x\in B(x_n,\epsilon_n)$ for any $n \in \mathbb{N}$, thus $x\in\underset{n\in\mathbb{N}}{\cap}B(x_n,\epsilon_n)$.

It is obvious from $B(x_n,\epsilon_n)\subseteq B(x_{n-1},\epsilon_{n-1}/2)\cap F_n^c$ that $B(x_n,\epsilon_n)\subseteq F_n^c$. Consequently, $x\in\underset{n\in\mathbb{N}}{\cap}B(x_n,\epsilon_n)\subseteq\underset{n\in\mathbb{N}}{\cap}F_n^c=\left(\underset{n\in\mathbb{N}}{\cup}F_n\right)^c=S^c$. The contradiction $x\in S^c=\emptyset$ has been reached, which completes the proof.

The case of finite union $S=\overset{k}{\underset{n=1}{\cup}}F_n$ is simpler, as it is not needed to construct an infinite sequence of open balls and then take the limit of their centers. Instead, it suffices to observe that the center $x_k$ of the smallest ball $B(x_k,\epsilon_k)$ satisfies $x_k\in\overset{k}{\underset{n=1}{\cap}}B(x_n,\epsilon_n)\subseteq\overset{k}{\underset{n=1}{\cap}} F_n^c$, leading to the analogous contradiction $x_k\in S^c=\emptyset$.

On a final note, it is emphasized that the existence of the countable sequence of ball centers $(x_n)$ is based on the axiom of choice.

# Clarification on the proof that an open set is a countable union of almost disjoint boxes

To establish that the Lebesgue outer measure of any open set in the Euclidean metric space of $\mathbb{R}^d$ is equal to the volume of any partitioning of that set into almost disjoint boxes, lemma 1.2.11 in p. 24 of the book “An introduction to measure theory” by Terence Tao first states that any open set $E\subseteq \mathbb{R}^d$ can be expressed as a countable union of almost disjoint boxes (and in fact as a countable union of almost disjoint closed cubes).

Note that this lemma is a generalization of the fact that every open subset of $\mathbb{R}$ can be expressed as a countable union of disjoint open intervals (see theorem 1.3, p. 6, in the book “Real analysis: measure theory, integration and Hilbert spaces” by Elias M. Stein and Rami Shakarchi and also lemma 2 of this blog post of mine).

The purpose of the present blog post is to elaborate on a detail in the proof of lemma 1.2.11 of Tao’s book. In particular, it will be explained why  every closed dyadic cube $Q$ is contained in exactly one maximal cube $Q^{*}$, without reproducing the rest of the proof.

Introduce the notation $Q_{n,\mathbf{i}}=\overset{d}{\underset{k=1}{\times}}\left[\frac{i_k}{2^n},\frac{i_k+1}{2^n}\right]$ to indicate the dependence of closed dyadic cubes on $n\in\mathbb{N}\cup \left\{0\right\}$ and $\mathbf{i}=(i_1,i_2,\cdots, i_d)\in\mathbb{Z}^{d}$. For every $\mathbf{i}=(i_1,i_2,\cdots, i_d)\in\mathbb{Z}^{d}$ for which there exists a closed dyadic cube $Q_{n,\mathbf{i}}$ contained in $E$, choose the biggest closed dyadic cube $Q_{n,\mathbf{i}}$ in $E$, i.e. choose $n_0=\underset{n}{\min}\left\{n\in\mathbb{N}\cup \left\{0\right\}:Q_{n,\mathbf{i}}\in E\right\}$. It is now obvious that by having capped the cubes by a sidelength of 1, there exists a maximum cube $Q_{n_0,\mathbf{i}}$ among $\left\{Q_{n,\mathbf{i}}\in E:n\in\mathbb{N}\cup \left\{0\right\}\right\}$, which is a maximal cube among those contained in $E$.

What is left to show is that every closed dyadic cube $Q$ can’t be contained in two or more maximal cubes. Assume that $Q$ is contained in two distinct maximal cubes $Q^*$ and $Q^{**}$, i.e. $Q\subseteq Q^*$ and $Q\subseteq Q^{**}$ with $Q^*\neq Q^{**}$. The dyadic nesting property then leads to the contradiction $Q^*\subseteq Q^{**}$ or $Q^{**}\subseteq Q^*$ for the maximal cubes $Q^*$ and $Q^{**}$, since $Q\subseteq Q^*$ and $Q\subseteq Q^{**}$ excludes the possibility of $Q^*$ and $Q^{**}$ being almost disjoint.

# Proof: the Hausdorff metric is finite under the boundedness assumption

Let $(E,\rho)$ be a metric space and $\mathcal{F}$ the family of non-empty, closed and bounded subsets of $E$. For any $X,Y\in\mathcal{F}$, define

$d(X, Y)=\max\{\sup\{\rho(x,Y):x\in X\},\sup\{\rho(y,X):y\in Y\}\},$

where $\rho(x,Y)$ denotes the distance of point $x\in X$ from set $Y$ and $\rho(y,X)$ is the distance of point $y\in Y$ from set $X$, that is

$\rho(x,Y)=\inf\{\rho(x,y):y\in Y\},~\rho(y,X)=\inf\{\rho(y,x):x\in X\}.$

It can be shown that $(\mathcal{F},d)$ is a metric space, and the metric $d$ is known as the Hausdorff metric. In order to prove that $d:\mathcal{F}\times\mathcal{F}\rightarrow \mathbb{R}$ is a metric, the following three properties of (the definition of) metric need to be shown:

1. $d(X,Y)=0\Leftrightarrow X=Y,$
2. $d(X,Y)=d(Y,X),$
3. $d(X,Y)\le d(X,Z)+d(Z,Y).$

Proofs of the validity of these three properties for the Hausdorff metric are readily available in several books of topology. To ensure that the Hausdorff metric is well-defined, its finiteness will be shown.

Let $x\in X$ and $y\in Y$. Then, since

$\rho(X,Y)=\inf\{\rho(x,y):(x,y)\in X\times Y\}$,

it follows that for every $n\in\mathbb{N}$ there exist $z\in X$ and $w\in Y$ such that

$\rho(z,w)\le\rho(X,Y)+\frac{1}{n}.\ \ \ \ (1)$

Indeed, assume that (1) is not true. Then there exists $n_0\in\mathbb{N}$ such that for all $z\in X$ and $w\in Y$ it holds that

$\rho(z,w)>\rho(X,Y)+\frac{1}{n_0},$

whence

$\inf\{\rho(z,w):(z,w)\in X\times Y\}=\rho(X,Y)\ge \rho(X,Y)+\frac{1}{n_0}\Rightarrow\frac{1}{n_0}\le 0,$

so a contradiction has been reached, which means that (1) actually holds.

According to the triangle inequality of metric $\rho$,

$\rho(x,y)\le \rho(x,z)+\rho(z,w)+\rho(w,y).\ \ \ \ (2)$

It thus follows from (1) and (2) that for every $n\in\mathbb{N}$

$\rho(x,y)\le \delta(X)+\rho(X,Y)+\delta(Y)+\frac{1}{n}.\ \ \ \ (3)$

Note that (3) holds because the sets $X,~Y$ are bounded, which means that their respective diameters $\delta(X),~\delta(Y)$ satisfy

$\rho(x,z)\le\delta(X)=\sup\{\rho(q,q):q\in X\}<\infty,$

$\rho(w,y)\le\delta(Y)=\sup\{\rho(q,q):q\in Y\}<\infty.$

Since (3) holds for all $n\in\mathbb{N}$, it follows that

$\rho(x,y)\le \delta(X)+\rho(X,Y)+\delta(Y).\ \ \ \ (4)$

Indeed, assume that (4) is not true. Then it holds that

$\delta(X)+\rho(X,Y)+\delta(Y)<\rho(x,y),$

and, in combination with (3), it is deduced that for any $n\in\mathbb{N}$

$0<\rho(x,y)-(\delta(X)+\rho(X,Y)+\delta(Y))\le\frac{1}{n},$

which is a contradiction because it follows from the Archimedean property that there exists $n_0\in\mathbb{N}$ such that

$\frac{1}{n_0}<\rho(x,y)-(\delta(X)+\rho(X,Y)+\delta(Y)),$

hence (4) is true.

Recalling that

$\rho(x,Y)=\inf\{\rho(x,y):y\in Y\}\le\rho(x,y),$

it follows from (4) that for any $x\in X$

$\rho(x,Y)\le \delta(X)+\rho(X,Y)+\delta(Y).\ \ \ \ (5)$

Since (5) holds for any $x\in X$, it follows that the set $\{\rho(x,Y):y\in Y\}$ has an upper bound, therefore according to the completeness axiom for real numbers, $\{\rho(x,Y):y\in Y\}$ has a supremum. In particular,

$\sup\{\rho(x,Y):x\in X\}\le \delta(X)+\rho(X,Y)+\delta(Y).\ \ \ \ (6)$

By means of symmetry of (6), this also means that

$\sup\{\rho(y,X):y\in Y\}\le \delta(X)+\rho(X,Y)+\delta(Y),\ \ \ \ (7)$

which completes the proof.